/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES After renaming modulo the bijection { a ↦ 0, d ↦ 1, b ↦ 2, c ↦ 3, f ↦ 4 }, it remains to prove termination of the 4-rule system { 0 1 ⟶ 1 2 3 2 1 , 0 ⟶ 2 2 4 2 2 , 2 1 2 ⟶ 0 1 , 1 4 ⟶ 2 1 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 3: 0 ↦ ⎛ ⎞ ⎜ 1 1 0 ⎟ ⎜ 0 1 0 ⎟ ⎜ 0 5 1 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 1 0 1 ⎟ ⎜ 0 1 0 ⎟ ⎜ 0 0 4 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 0 0 ⎟ ⎜ 0 1 0 ⎟ ⎜ 0 1 1 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 0 0 ⎟ ⎜ 0 1 0 ⎟ ⎜ 0 0 0 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 1 0 0 ⎟ ⎜ 0 1 0 ⎟ ⎜ 0 1 1 ⎟ ⎝ ⎠ After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2, 3 ↦ 3 }, it remains to prove termination of the 2-rule system { 0 1 ⟶ 1 2 3 2 1 , 2 1 2 ⟶ 0 1 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 4: 0 ↦ ⎛ ⎞ ⎜ 1 0 1 0 ⎟ ⎜ 0 1 0 0 ⎟ ⎜ 0 1 0 0 ⎟ ⎜ 0 0 0 0 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 1 0 0 0 ⎟ ⎜ 0 1 0 0 ⎟ ⎜ 0 0 0 1 ⎟ ⎜ 0 0 0 0 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 0 1 1 ⎟ ⎜ 0 1 0 0 ⎟ ⎜ 0 1 0 0 ⎟ ⎜ 0 1 0 0 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 0 0 0 ⎟ ⎜ 0 1 0 0 ⎟ ⎜ 0 0 0 0 ⎟ ⎜ 0 0 0 0 ⎟ ⎝ ⎠ After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2, 3 ↦ 3 }, it remains to prove termination of the 1-rule system { 0 1 ⟶ 1 2 3 2 1 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 2: 0 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ After renaming modulo the bijection { }, it remains to prove termination of the 0-rule system { } The system is trivially terminating.